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Is sum of torque zero? What are the possible errors in the lab?

Is sum of torque zero? What are the possible errors in the lab?

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Step 5: Find the distance r and rz from r., record data in observation table below. Example: In figure below, r2 =- 20 + 50 = 30 cm, r = 87 - 50 = 37 cm 60 100 Fz=mag Styrofoam cup F=mag Find force Find force F2of the F of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F;=m.g and F=mag Step 7: Exchange weights m, and m2 and repeat from step 4 Step 8: There is no step 8 Distance from Torque 1= Fr Sum of F= mg (N) g= 9.8 m/s? Mass (m) in Kg Obs. # center of ruler Torques i= 11+ 12 (1g=0.001kg) (1cm = 0.01m) F T= Fn 0.05 1 F2 m2 0.10 F1 0.10 2 F2 m2 0.05 12 = - m Is sum of torque zero? What are the possible errors in the lab?
Transcribed Image Text:Step 5: Find the distance r and rz from r., record data in observation table below. Example: In figure below, r2 =- 20 + 50 = 30 cm, r = 87 - 50 = 37 cm 60 100 Fz=mag Styrofoam cup F=mag Find force Find force F2of the F of the hanging hanging mass mass Fig 1: Experimental set up Step 6: Find Force on each end of the ruler by substituting vales for m, and m2 in equation F;=m.g and F=mag Step 7: Exchange weights m, and m2 and repeat from step 4 Step 8: There is no step 8 Distance from Torque 1= Fr Sum of F= mg (N) g= 9.8 m/s? Mass (m) in Kg Obs. # center of ruler Torques i= 11+ 12 (1g=0.001kg) (1cm = 0.01m) F T= Fn 0.05 1 F2 m2 0.10 F1 0.10 2 F2 m2 0.05 12 = - m Is sum of torque zero? What are the possible errors in the lab?
Static Equilibrium of Rigid Bodies: Torques Goals Study the relationship between force, lever arm, and torque; study how torques add Study the conditions for static equilibrium of a rigid body Study the concept of “center of gravity" Introduction and Background Torques: The torque rcreated by a force F with respect to a certain axis of rotation is 1 = Fx R1 where R is the so-called lever arm, which is the perpendicular (shortest) distance between the axis (6-1) of rotation and the line of force. Torques are vectors. Strictly speaking, the direction of a torque should be perpendicular to (either into or out of) the plane defined by the force and its lever arm. However, for simplicity, we will define the direction of a torque as either clockwise (CW) or counterclockwise (CCW) depending which way the torque would make the object rotate about the axis of rotation chosen. Either CW or CCW can be chosen as the "+" direction as long as consistency is maintained. We will choose CW as the +" direction here. The resultant of multiple torques can then be obtained as the algebraic sum of the individual torques. Static Equilibrium: In order for a rigid body to stay in equilibrium, the net force and the net torque on the object must both be zero. We usually only deal with motion of objects in two dimensions, in which case the conditions for static equilibrium are EF. = 0, EF, = 0 (6-2) Στ 0 (6-3) The second condition should hold for any choice of the axis of rotation. Center of Gravity: For a rigid body with finite size, the force of gravity acts on all parts of the body. But for the purpose of studying the translational motion of the body as a whole or the static equilibrium of the body, we can assume that the entire weight of the body acts at a single point. This point is the center of gravity. When we draw the force diagram for a rigid body, we can put a single force of gravity at the center of gravity. Experimental Setup Equipment: Meter stick, fulcrum, clamps, weights, pan balance r2 Weight 1 (50gm) Weight 2 Fuldrum (100gm) Setup: The experimental setup schematically in Figure is 15 shown Figure 1- Arrangement for balancing toraues Fig 1. The apparatus consists of a meter stick suspended from a fulcrum. Weights may be hung from the meter stick at various positions along the stick by means of special clamps in order to apply torques. Experimental Procedure and Data Analysis A. Torques 1. Adjust the location of the fulcrum so that the meter stick, with no weights hanging on it, is in static equilibrium (balance) in a horizontal position. The position of the fulcrum is roughly the center of gravity of the meter stick. Note that this position is not necessarily in the middle of the stick (50 cm mark). 2. With a clamp, hang a 50-g mass on the stick 20 cm from the fulcrum. Hang another 100-g mass with a clamp on the opposite side of the stick. Move one of the mass and find a position so that the meter stick is again in equilibrium in a horizontal position.
Transcribed Image Text:Static Equilibrium of Rigid Bodies: Torques Goals Study the relationship between force, lever arm, and torque; study how torques add Study the conditions for static equilibrium of a rigid body Study the concept of “center of gravity" Introduction and Background Torques: The torque rcreated by a force F with respect to a certain axis of rotation is 1 = Fx R1 where R is the so-called lever arm, which is the perpendicular (shortest) distance between the axis (6-1) of rotation and the line of force. Torques are vectors. Strictly speaking, the direction of a torque should be perpendicular to (either into or out of) the plane defined by the force and its lever arm. However, for simplicity, we will define the direction of a torque as either clockwise (CW) or counterclockwise (CCW) depending which way the torque would make the object rotate about the axis of rotation chosen. Either CW or CCW can be chosen as the "+" direction as long as consistency is maintained. We will choose CW as the +" direction here. The resultant of multiple torques can then be obtained as the algebraic sum of the individual torques. Static Equilibrium: In order for a rigid body to stay in equilibrium, the net force and the net torque on the object must both be zero. We usually only deal with motion of objects in two dimensions, in which case the conditions for static equilibrium are EF. = 0, EF, = 0 (6-2) Στ 0 (6-3) The second condition should hold for any choice of the axis of rotation. Center of Gravity: For a rigid body with finite size, the force of gravity acts on all parts of the body. But for the purpose of studying the translational motion of the body as a whole or the static equilibrium of the body, we can assume that the entire weight of the body acts at a single point. This point is the center of gravity. When we draw the force diagram for a rigid body, we can put a single force of gravity at the center of gravity. Experimental Setup Equipment: Meter stick, fulcrum, clamps, weights, pan balance r2 Weight 1 (50gm) Weight 2 Fuldrum (100gm) Setup: The experimental setup schematically in Figure is 15 shown Figure 1- Arrangement for balancing toraues Fig 1. The apparatus consists of a meter stick suspended from a fulcrum. Weights may be hung from the meter stick at various positions along the stick by means of special clamps in order to apply torques. Experimental Procedure and Data Analysis A. Torques 1. Adjust the location of the fulcrum so that the meter stick, with no weights hanging on it, is in static equilibrium (balance) in a horizontal position. The position of the fulcrum is roughly the center of gravity of the meter stick. Note that this position is not necessarily in the middle of the stick (50 cm mark). 2. With a clamp, hang a 50-g mass on the stick 20 cm from the fulcrum. Hang another 100-g mass with a clamp on the opposite side of the stick. Move one of the mass and find a position so that the meter stick is again in equilibrium in a horizontal position.
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